Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define fill( x, y ) memset( x, y, sizeof x )
using namespace std;

typedef long long LL;
typedef pair < int, int > pa;

const int MAXN = 100010;
const LL INF = 1e18;

int n, K, A[MAXN], L[MAXN], R[MAXN];
int ls[MAXN << 1], tot, type[MAXN], e[MAXN << 2], sum;
LL f[MAXN], g[MAXN], ans = INF;

inline void modify(int x, int l, int r, int ql, int qr, int v)
{
	if( l == ql && r == qr ) { e[ x ] = v; return ; }
	int mid = l + r >> 1;
	if( qr <= mid ) modify( x << 1, l, mid, ql, qr, v );
	else if( ql > mid ) modify( x << 1 | 1, mid + 1, r, ql, qr, v );
	else modify( x << 1, l, mid, ql, mid, v ), modify( x << 1 | 1, mid + 1, r, mid + 1, qr, v );
}

inline int querymin(int x, int l, int r, int v)
{
	if( l == r ) return e[ x ];
	int mid = l + r >> 1;
	if( v <= mid ) return min( e[ x ], querymin( x << 1, l, mid, v ) );
	return min( e[ x ], querymin( x << 1 | 1, mid + 1, r, v ) );
}

inline void add(int x, int l, int r, int ql, int qr)
{
	if( l == ql && r == qr ) { e[ x ]++; return ; }
	int mid = l + r >> 1;
	if( qr <= mid ) add( x << 1, l, mid, ql, qr );
	else if( ql > mid ) add( x << 1 | 1, mid + 1, r, ql, qr );
	else add( x << 1, l, mid, ql, mid ), add( x << 1 | 1, mid + 1, r, mid + 1, qr );
}

inline int querysum(int x, int l, int r, int v)
{
	if( l == r ) return e[ x ];
	int mid = l + r >> 1;
	if( v <= mid ) return e[ x ] + querysum( x << 1, l, mid, v );
	return e[ x ] + querysum( x << 1 | 1, mid + 1, r, v );
}

int main()
{
#ifdef wxh010910
	freopen( "data.in", "r", stdin );
	freopen( "my.out", "w", stdout );
#endif
	scanf( "%d%d", &n, &K );
	for( int i = 1 ; i <= n ; i++ )
	{
		scanf( "%d%d", &A[ i ], &type[ i ] );
		if( A[ i ] * 2 > K && type[ i ] == 1 ) return printf( "-1\n" ), 0;
		if( type[ i ] == 2 ) L[ i ] = 0, R[ i ] = K - 1;
		else L[ i ] = sum, R[ i ] = ( ( L[ i ] - 2 * A[ i ] ) % K + K ) % K;
		sum = ( ( sum - 2 * A[ i ] ) % K + K ) % K;
	}
	for( int i = 1 ; i <= n ; i++ ) ls[ ++tot ] = L[ i ], ls[ ++tot ] = R[ i ];
	sort( ls + 1, ls + tot + 1 ); tot = unique( ls + 1, ls + tot + 1 ) - ls - 1;
	for( int i = 1 ; i <= n ; i++ ) L[ i ] = lower_bound( ls + 1, ls + tot + 1, L[ i ] ) - ls, R[ i ] = lower_bound( ls + 1, ls + tot + 1, R[ i ] ) - ls;
	for( int i = 1 ; i <= ( n << 2 ) ; i++ ) e[ i ] = n + 1;
	for( int i = n, pos ; i ; i-- )
	{
		pos = querymin( 1, 1, tot, L[ i ] );
		if( pos != n + 1 ) f[ i ] = f[ pos ] + ( ls[ L[ pos ] ] - ls[ L[ i ] ] + K ) % K;
		pos = querymin( 1, 1, tot, R[ i ] );
		if( pos != n + 1 ) g[ i ] = f[ pos ] + ( ls[ L[ pos ] ] - ls[ R[ i ] ] + K ) % K;
		if( L[ i ] <= R[ i ] )
		{
			if( R[ i ] < tot ) modify( 1, 1, tot, R[ i ] + 1, tot, i );
			if( L[ i ] > 1 ) modify( 1, 1, tot, 1, L[ i ] - 1, i );
		}
		else if( R[ i ] + 1 < L[ i ] ) modify( 1, 1, tot, R[ i ] + 1, L[ i ] - 1, i );
	}
	for( int i = 1 ; i <= ( n << 2 ) ; i++ ) e[ i ] = 0;
	for( int i = 1 ; i <= n ; i++ )
	{
		if( querysum( 1, 1, tot, L[ i ] ) == i - 1 )
			ans = min( ans, f[ i ] );
		if( querysum( 1, 1, tot, R[ i ] ) == i - 1 )
			ans = min( ans, g[ i ] );
		if( L[ i ] <= R[ i ] ) add( 1, 1, tot, L[ i ], R[ i ] );
		else add( 1, 1, tot, L[ i ], tot ), add( 1, 1, tot, 1, R[ i ] );
	}
	for( int i = 1 ; i <= n ; i++ ) ans += A[ i ] << 1;
	cout << ans << endl;
}